Let $g(x)=x^4-x^5$. For what value of $x$ does $g$ have a relative maximum ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{4}{5}$ (Choice B) B $-\dfrac{5}{4}$ (Choice C) C $1$ (Choice D) D $0$
Explanation: We can find the relative extrema (i.e. minima and maxima) of $g$ by looking for the intervals where its derivative $g'$ is positive/negative. A function can only change its direction from increasing to decreasing and vice versa between its critical points and the points where the function itself is undefined. The derivative of $g$ is $g'(x)=x^3(4-5x)$. $g'(x)=0$ for $x=0,\dfrac{4}{5}$. Since $g'$ is a polynomial, it's defined for all real numbers. Therefore, our critical points are $x=0$ and $x=\dfrac{4}{5}$. $g$ is defined for all real numbers so we only need to consider the critical points. Our critical points divide the number line into three intervals: $\llap{-}2$ $\llap{-}1$ $0$ $1$ $2$ $3$ $4$ $x<0$ $0<x<\frac{4}{5}$ $\frac{4}{5}$ $x>\frac{4}{5}$ Let's evaluate $g'$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $g'(x)$ Verdict $x<0$ $x=-1$ $g'(-1)=-9<0$ $g$ is decreasing $\searrow$ $0<x<\dfrac{4}{5}$ $x=\dfrac15$ $g'\left(\dfrac15\right)=\dfrac{3}{125}>0$ $g$ is increasing $\nearrow$ $x>\dfrac{4}{5}$ $x=1$ $g'(1)=-1<0$ $g$ is decreasing $\searrow$ Now let's look at the critical points: $x$ Before After Verdict $0$ $\searrow$ $\nearrow$ Minimum $\dfrac{4}{5}$ $\nearrow$ $\searrow$ Maximum Now we can see that $g$ has a relative maximum at $x=\dfrac{4}{5}$.